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Theorem merlem3 1409
 Description: Step 7 of Meredith's proof of Lukasiewicz axioms from his sole axiom. (Contributed by NM, 14-Dec-2002.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
merlem3 (((ψχ) → φ) → (χφ))

Proof of Theorem merlem3
StepHypRef Expression
1 merlem2 1408 . . . 4 (((¬ χ → ¬ χ) → (¬ χ → ¬ χ)) → ((φφ) → (¬ χ → ¬ χ)))
2 merlem2 1408 . . . 4 ((((¬ χ → ¬ χ) → (¬ χ → ¬ χ)) → ((φφ) → (¬ χ → ¬ χ))) → ((((χφ) → (¬ ψ → ¬ ψ)) → ψ) → ((φφ) → (¬ χ → ¬ χ))))
31, 2ax-mp 8 . . 3 ((((χφ) → (¬ ψ → ¬ ψ)) → ψ) → ((φφ) → (¬ χ → ¬ χ)))
4 ax-meredith 1406 . . 3 (((((χφ) → (¬ ψ → ¬ ψ)) → ψ) → ((φφ) → (¬ χ → ¬ χ))) → ((((φφ) → (¬ χ → ¬ χ)) → χ) → (ψχ)))
53, 4ax-mp 8 . 2 ((((φφ) → (¬ χ → ¬ χ)) → χ) → (ψχ))
6 ax-meredith 1406 . 2 (((((φφ) → (¬ χ → ¬ χ)) → χ) → (ψχ)) → (((ψχ) → φ) → (χφ)))
75, 6ax-mp 8 1 (((ψχ) → φ) → (χφ))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4 This theorem was proved from axioms:  ax-mp 8  ax-meredith 1406 This theorem is referenced by:  merlem4  1410  merlem6  1412
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